Let $m$ be an integer and $n$ a positive factor of $m$ . If $x$ is a positive real number, we may write the identical equation $$(x^{\frac{m}{n}})^n = x^{\frac{m}{n}\cdot n} = x^m$$ and therefore the definition of $n^\mathrm{th}$ root gives the formula

(1)

Definition. Let $\frac{m}{n}$ be a fractional number, i.e. an integer $m$ not divisible by the integer $n$ , which latter we assume to be positive. For any positive real number $x$ we define the fraction power $x^{\frac{m}{n}}$ as the $n^\mathrm{th}$ root

(2)

1. The existence of the root in the right hand side of (2) is proved here.
2. The defining equation (2) is independent on the form of the exponent $\frac{m}{n}$ : If $\frac{k}{l} = \frac{m}{n}$ , then we have $(\sqrt[n]{x^m})^{ln} = [(\sqrt[n]{x^m})^n]^l = x^{lm} = x^{kn} = [(\sqrt[l]{x^k})^l]^n = (\sqrt[l]{x^k})^{ln}$ , and because the mapping $y\mapsto y^{ln}$ is injective in $\mathbb{R}_+$ , the positive numbers $\sqrt[l]{x^k}$ and $\sqrt[n]{x^m}$ must be equal.
3. The fraction power function $x\mapsto x^{\frac{m}{n}}$ is a special case of power function.
4. The presumption that $x$ is positive signifies that one can not identify all $n^\mathrm{th}$ roots $\sqrt[n]{x}$ and the powers $x^{\frac{1}{n}}$ . For example, $\sqrt[3]{-8}$ equals $-2$ and $\frac{2}{6} = \frac{1}{3}$ , but one must not calculate $$(-8)^{\frac{1}{3}} = (-8)^{\frac{2}{6}} = \sqrt[6]{(-8)^2} = \sqrt[6]{64} = 2.$$ The point is that $(-8)^{\frac{1}{3}}$ is not defined in $\mathbb{R}$ . Here we have $l = 6$ and the mapping $y\mapsto y^{ln}$ is not injective in $\mathbb{R}_-\cup\mathbb{R}_+$ . -- Nevertheless, some people and books may use also for negative $x$ the equality $\sqrt[n]{x} = x^{\frac{1}{n}}$ and more generally $\sqrt[n]{x^m} = x^{\frac{m}{n}}$ where one then insists that $\gcd(m,\,n) = 1.$
5. According to the preceding item, for the negative values of $x$ the derivative of odd roots, e.g. $\sqrt[3]{x}$ , ought to be calculated as follows: $$\frac{d\sqrt[3]{x}}{dx} = \frac{d(-\sqrt[3]{-x})}{dx} = -\frac{d(-x)^\frac{1}{3}}{dx} = -\frac{1}{3}\cdot(-x)^{-\frac{2}{3}}(-1) = \frac{1}{3\sqrt[3]{(-x)^2}} = \frac{1}{3\sqrt[3]{x^2}}$$ The result is similar as $\frac{d\sqrt[3]{x}}{dx}$ for positive $x$ 's, although the odd root functions are not special cases of the power function.